Quiz

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Parametric Arclength is the length of a curve given by parametric equations. For instance, the curve in the image to the right is the graph of the parametric equations \(x(t) = t^2 + t\) and \(y(t) = 2t - 1\) with the parameter \(t\). One could wish to find the arclength of curve between the points \(t =-\frac{1}{2}\) and \(t=1\), as noted by the thicker red curve in the image to the right.

Generalized, a parametric arclength starts with a parametric curve in \(\mathbb{R}^2\). This is given by some parametric equations \(x(t)\), \(y(t)\), where the parameter \(t\) ranges over some given interval. The following formula computes the length of the arc between two points \(a,b\).

Consider a parametric curve \((x(t), y(t))\), where \(t\in [a,b]\). The length of the arc traced by the curve as \(t\) ranges over \([a,b]\) is \[ \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2} \, dt .\]

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Derivation Of Formula

Consider the curve given by parametric equations \(x(t)\), \(y(t)\), where \(t\) ranges over \([0,1]\). To compute the arclength of this curve, one can approximate the curve with polygonal paths and take the limit of the lengths of these paths as they become more refined.

A curve in the plane, approximated a polygonal path consisting of straight line segments.

In particular, if \(f: \mathbb{R} \to \mathbb{R}\) is differentiable, the graph of \(f\) can be parametrized as \(t \mapsto (t, f(t))\). Then, the arclength of the segment of this graph bounded by \((a, f(a))\) and \((b,f(b))\) equals \[\int_{a}^{b} \sqrt{1+(f'(t))^2} \, dt.\]

Let \(\Gamma\) denote the graph of \(f(x) = \ln(\sec(x))\). What is the arclength of the segment of \(\Gamma\) bounded by the points \((0,0)\) and \( \left (\dfrac \pi4, \ln(\sqrt{2})\right)\)?

Surface Area

Suppose the graph of \(y = f(x)\) is rotated about the \(x\)-axis to obtain a solid. By the above formula for arclength, the infinitesimal arclength of this graph at some \(x\in \mathbb{R}\) is \(\sqrt{1+(f'(x))^2} \, dx\). When this arc of the graph is rotated about the \(x\)-axis, the solid formed is an infinitesimal cylinder of surface area \(2\pi f(x) \sqrt{1+(f'(x))^2} \, dx\). Integrating this over the desired interval, one obtains the formula below.

Let \(\gamma\) denote the arc of the graph \(y = f(x)\) bounded by the points \((a,f(a))\) and \((b,f(b))\). The surface area of the solid formed by rotating \(\gamma\) about the \(x\)-axis is \[2\pi \int_{a}^{b} f(x) \sqrt{1+(f'(x))^2} \, dx.\]